The minimum number of condensers each capacitance of `2 muF`, in order to obtain result capacitance of `5 muF` will be

Options

A. 4
B. 5
C. 6
D. 3

Answer

To solve the problem of finding the minimum number of capacitors, each with a capacitance of \(2 \mu F\), required to achieve a resultant capacitance of \(5 \mu F\), we can follow these steps:

### Step 1: Understand the Capacitor Combinations
We can combine capacitors in two ways: in series and in parallel. The total capacitance for capacitors in series is given by:
\[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots \]
For capacitors in parallel, the total capacitance is simply the sum of the individual capacitances:
\[ C_{total} = C_1 + C_2 + \ldots \]

### Step 2: Calculate the Capacitance in Series
To achieve a capacitance of \(5 \mu F\), we can first consider creating a smaller capacitance using the series combination. If we connect two \(2 \mu F\) capacitors in series:
\[ \frac{1}{C_{series}} = \frac{1}{2} + \frac{1}{2} = \frac{1}{1} \Rightarrow C_{series} = 1 \mu F \]

### Step 3: Create the Required Capacitance
Now, we need to achieve \(5 \mu F\). We can do this by combining the \(1 \mu F\) capacitor (from the series combination) with other \(2 \mu F\) capacitors in parallel.

If we connect two \(2 \mu F\) capacitors in parallel:
\[ C_{parallel} = 2 + 2 = 4 \mu F \]

Now, we can combine the \(1 \mu F\) capacitor (from the series combination) with the \(4 \mu F\) capacitor (from the parallel combination):
\[ C_{total} = 1 + 4 = 5 \mu F \]

### Step 4: Count the Total Number of Capacitors Used
- **For the series combination**: We used **2 capacitors** to create \(1 \mu F\).
- **For the parallel combination**: We used **2 capacitors** to create \(4 \mu F\).

Thus, the total number of capacitors used is:
\[ 2 + 2 = 4 \]

### Conclusion
The minimum number of \(2 \mu F\) capacitors required to achieve a resultant capacitance of \(5 \mu F\) is **4 capacitors**.